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1 + 10 square theta divided by 1 + cot square theta is equal to 1 minus 10 theta divided by 1 minus cot theta whole square |
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Answer» Step-by-step explanation: The given equation is: \FRAC{1+tan^2{\theta}}{1+cot^2{theta}}=(\frac{1-tan{\theta}}{1-cot{\theta}})^{2} TAKING the LHS of the above equation, we have \frac{1+tan^2{\theta}}{1+cot^2{theta}} =\frac{sec^2\theta}{cosec^2\theta} =\frac{sin^2\theta}{cos^2\theta} =tan^2{\theta} Now, taking the RHS of the above equation, we have (\frac{1-tan{\theta}}{1-cot{\theta}})^{2} =(\frac{1-tan{\theta}}{1-\frac{1}{tan\theta}})^{2} =(\frac{(1-tan{\theta})tan{\theta}}{-(1-tan{\theta})})^2 =tan^2{\theta} Hence, LHS=RHS, THUS proved. 4.0 |
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